MATLAB. HELP! I DONT UNDERSTAND FIXED POINT ITERATION AT ALL!!

L. Implement fixed-point iteration. Your signature should be function p fp(g, po, naxita) where g is a function handle and po, maxits, and p are numbers. The return value p is the result of maxita applications of g, e p = g(g( . . . g(po)--)). Test your code on the following four functions. (a) g: (z +2/a)/2. (This is the .Babylonian method, for computing V2 = 1.414-.) . Contir visually that the fixed-point problem "Solve g(r) "is equivalent to the root-finding problem *Solve f(z-0, with f: z zs_ 2. That is, make a plot showing that the same r-values which y-) intersects y-a. = f(z) intersects = 0 at pO and naxits. How can you pick pO to control which of the two iii. Try po = 1.5 and compare with the bisection method (Worksheet 3) with a = 1,b = 2, and tol = epe ii. Experiment with the fixed points (tv2) the method converges to? Plot the eror (abs (aqrt (2) p)) of the two methods as a function of iteration number (You will have to modify your fp and bisection codes to return all of the iterates, not just the last.) Use a senilogy plot. (b) g: cos() . Plot y-g() and y-z to visually confirm that g has a unique fixed point. ii. Plot the error as a function of iteration number, using the converged value as the exact solution. ii. (Challenge: unguded) Estimate the rate at which the error (as a function of iteration number n converges (to zero) when po 1. Do this by fitting a curve of the form nH C" to the error (c) g:r 2r. This function has a unique fixed point at z = 0, For what starting value(s) po does fixed point iteration converge to it? What happens otherwise This function has no fixed points. What does your fp implementation do? If you performed fixed-point iteration in exact (infinite precision) arithmetie, what would happen instead